【K12教育学习资料】[学习]重庆市外国语学校(四川外国语大学附属外国语学校)2018-2019学年

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重庆市外国语学校四川外国语大学附属外国语学校高二英语上学期半期考试试题

重庆市外国语学校四川外国语大学附属外国语学校高二英语上学期半期考试试题

重庆外国语学校 2021-2021学年〔上〕高二年级半期考试一试题英语考前须知:1.本试卷分第一卷〔选择题,1至10页〕和第二卷〔非选择题,11至12页〕两局部;考试时间120分钟;总分值130分。

所有答案在答题卡上达成,答在本试卷上无效。

答题前,考生务势必自己的姓名、准考据号等填涂在答题卡指定的地点。

第一局部听力〔共两节,总分值20分〕第一节听下边5段对话,每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最正确选项。

听完每段对话后,你都有10秒钟的时间往返答相关小题和阅读下一小题。

每段对话仅读一遍。

Whatdidthemanprobablyplantodo?WatchTV.Playfootball.Gotothecinema.Whatdidthewomanwantonthesideatfirst?A. Salad.B.Potatoes.C.Greenbeans.Whomightthewomanbeworriedabout?A. Herbrother.B.Herclassmate.C.Hersister.Whenwillthespeakersprobablywalkthedog?Inonehour.Inthirtyminutes.Infifteenminutes.Whydidthewomancutherhair?A.Itlookedbad.Ittooklongtodry.第-1- 页C. Itwasalwaysintheway.第二节听下边5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最正确选项。

听每段对话或独白前,你将有时间阅读各个小题,每题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段资料,答复第6至7题。

HowmuchmoneywillKlayThompsondonate?A. $29,000.B. $2900.$1000.Whatdoesthemanmeanattheend?Hewillalsodonatesomemoney.HewisheshewereasrichasLlayThompson.Heisn’timpressedbyarichpersondonatingmoney.听第7段资料,答复第8至10题。

重庆市外国语学校(四川外国语大学附属外国语学校)2018-2019学年高一历史上学期半期考试试题

重庆市外国语学校(四川外国语大学附属外国语学校)2018-2019学年高一历史上学期半期考试试题

重庆外国语学校2018—2019学年度上期期中考试高一历史试题(满分100分,时间100分钟)注意事项:1.答题前,考生务必将自己的班级、姓名、顺序号、准考证号用0.5毫米黑色签字笔填写在答题卡上,准考证号码还需用2B铅笔准确涂写。

2.每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑。

如需改动,用橡皮擦干净后,再选涂其它答案,不能答在试题卷上。

3.请认真书写,规范答题。

一、选择题(本大题共40小题,每小题1.5分,共60分)在每题给出的四个选项中,只有一项是最符合题目要求的。

1.据史载,夏后启讨伐有扈氏时说自己是“恭行天罚”,对部下们则说:“用命,赏于祖;不用命,戳于社”。

商汤伐夏时说:“有夏多罪,天命殛之”,“子畏上帝,不敢不征”。

周武王伐纣时也曾说:“商罪贯盈,天命诛之”。

该材料表明夏、商、周时期的中国( ) A.集权政治制度已形成 B.占卜和迷信活动盛行C.神权与王权有机结合 D.宗法等级观日渐凸显2.“封藩建卫”是在土地王有的前提下,利用宗法血缘纽带,“封建亲戚,以藩屏周”,各方诸侯都以周天子为天下共主,实行层层分封,形成了宗法制统治的奴隶制大国。

据此判断以下说法错误的是( )A.稳定了周王朝的政治秩序 B.宗法制是分封制的内在依据C.巩固和扩大了周王朝的统治D.有利于周王权力的高度集中3.周公把商贵族微子启分封到宋,同时又在宋周围分封同姓诸侯。

周公这样做的目的是( )A.与商贵族共掌天下 B.拉拢并控制商贵族C.巩固中央集权体制 D.确立自己的专制权威4.《史记》叙述西周历史时写道:“封周公旦于少昊之虚曲阜。

”下列有关周公受封的叙述,正确的是( )①按照血缘宗族关系分配权力②对燕国进行世袭统治③对天子定期朝觐并交纳各种贡赋④官僚政治进一步完善A.①②B.①③C.②③D.③④5.中国古代视太子为国本。

明朝神宗时期,先是恭妃生皇长子朱常洛,后来深受神宗宠爱的郑贵妃生皇三子朱常洵,神宗企图立朱常洵为太子,遭到朝臣坚决抵制。

【英语】重庆市外国语学校(四川外国语大学附属外国语学校)2018-2019学年高一上学期半期考试试题

【英语】重庆市外国语学校(四川外国语大学附属外国语学校)2018-2019学年高一上学期半期考试试题

重庆市外国语学校(四川外国语大学附属外国语学校)2018-2019学年高一上学期半期考试英语试题考试时间120分钟;满分130分。

第Ⅰ卷选择题部分第一部分听力(共两节,满分20分)第一节(共5小题;每小题1分,满分5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.What happened to the woman?A.She ate something strange.B.She missed a lunch meeting.C.She forgot about the dinner event.2.Where will the woman go for spring break?A.To Hawaii.B. To Texas.C. To Alaska3.What will the woman do next?A.Clean up her mess.B.Buy the flower pot.C.Pay for something she broke.4.Which dog does the man want?A.An adult male.B. A female puppy.C. An adult female.5.Why is the man so upset?A.His book is damaged.B.Someone checked out the book he wants.C.The woman won’t let him borrow the book.第二节(共15小题;每小题1分,满分15分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

重庆市外国语学校四川外国语大学附属外国语学校2018_2019学年高二历史上学期半期考试试题

重庆市外国语学校四川外国语大学附属外国语学校2018_2019学年高二历史上学期半期考试试题

重庆外国语学校2018—2019学年度(上)半期考试历史试题(满分100分,时间100分钟)一、选择题(本大题共30小题,每小题2分,共60分。

在每题给出的四个选项中,只有一项是最符合题目要求的。

)1.魏征在《谏太宗十思疏》中说“文武并用,垂拱而治。

何必劳神苦思,代百司之职役哉?下列与魏征这一主张最为相近的思想是A.老子无为而治思想B.韩非中央集权思想C.孟子民贵君轻思想D.孔子为政以德思想2.一般来说,当人对社会变迁感到难以适应时,可能采取不同的反应方式:(1)利用原有的行为规范。

(2)提出自己的行为方式、并努力使它为社会所采用。

(3)用各种反社会的言论和行为,以攻击现存的社会秩序。

孔子采取的方式是A.(1) B.(2)C.(1)(2) D.(1)(2)(3)3.“天之生民,非为君也,天下立君,以为民也。

故古者列地建国,非以贵诸侯而已;列官职,爵禄,非以尊大夫而已。

”在这里,荀子强调A.尊君 B.礼法C.重民 D.仁义4.张荫麟在《中国史纲》中说:“墨子以为天下的大害,莫如大国之侵略小国,大家族之欺凌小家族,强者智者之压迫弱者愚者,以及一切伦常间的失欢失德,总而言之,即人与人的冲突。

”为了消除“天下之大害”,墨子提倡A.仁爱 B.仁政C.尚贤 D.兼爱5.央视播出的成语大会中出现了成语“越俎代庖、舍生取义、守株待兔”。

这三个成语都与我国古代思想流派有关,分别是A.道家、儒家、法家B.墨家、儒家、法家C.兵家、墨家、道家D.儒家、道家、墨家6. 下面是儒家思想在古代的发展情况示意图。

下列对a、b、c、d四处出现起伏的原因的分析正确的是A. a处处于低潮受秦“罢黜百家,独尊儒术”思想的影响B. b处处于高潮是受汉朝焚书坑儒的推动C. c处处于低潮是受魏晋南北朝时期佛、道思想的冲击D. d处处于高潮是受明清时期批判思想的推动7. “大一统的封建帝国需要这样一种宗教:用统一的神权来神化君权,用祖先崇拜来巩固宗法等级制度,用‘天定’的禁欲主义清规戒律来束缚民众,用教主崇拜来管制思想……经历两次改造,儒家变为儒教,孔子抬高为被崇拜的偶像。

重庆市四川外国语大学附属外国语学校2024-2025学年高二上学期第一次月考英语试题

重庆市四川外国语大学附属外国语学校2024-2025学年高二上学期第一次月考英语试题

重庆市四川外国语大学附属外国语学校2024-2025学年高二上学期第一次月考英语试题一、阅读理解About Scholastic Kids PressScholastic Kids Press is a group of talented kid reporters, ages 10 — 14, from across the country and around the world. Since 2000, our award-winning young journalists have reported “news for kids, by kids”, covering politics, entertainment, the environment, sports and more in their hometowns and on the national stage. Their stories appear online and in Scholastic magazines, which reach more than 25 million students in classrooms nationwide.Scholastic Kid Reporters have made headlines by interviewing Pulitzer Prize-winning composer Lin-Manuel Miranda, environmentalist Jane Goodall, Librarian of Congress Carla Hayden, Nobel Peace Prize winner Malala Yousafzai, Dog Man author Dav Pilkey, actress Misty Copeland and more.Contact InfoEmail us at ************************.How to apply?We are now accepting Kid Reporter applications for the 2023-2024 program year. Learn more and download the application here.The application to join Scholastic Kids Press includes writing a news story, story ideas and a personal article. Editors select Kid Reporters every year based on writing ability, interviewing skills and attention to detail.To be considered, applicants must be between 10-14. Students in any part of the world are welcome to apply.What do Scholastic Kid Reporters write about?Throughout the year, Scholastic Kid Reporters cover issues and topics that matter most to them and their peers. This includes current events, breaking news, entertainment, sports and more from their communities.The Kid Reporter application is now open! Click here to learn more.1.What do we know about Scholastic Kids Press?A.It enjoys great popularity among adults.B.It is the press composed of young reporters.C.It publishes articles only in Scholastic magazines.D.It has an enormous influence on extraordinary people.2.Which of the requirements should an applicant meet?A.The place they come from.B.Their age.C.Their personal experience.D.The language they speak.3.What do Scholastic Kid Reporters mainly cover?A.Skills in making delicious food.B.Stories of their parents.C.The most important things for kids.D.Housework suitable for children.Most people who ran the marathon last month spent days before resting up and days after recovering. But not Jacky Hunt-Broersma. In total, she ran 104 marathons in 104 days. If it’s confirmed by Guinness World Records, that would be a new world record. And she did it all with one leg and a prosthesis (假肢).“Someone able-bodied had done it previously. I wanted to see if I could do it on a prosthetic leg and see what would happen,” she said. “I thought it would be a great way to inspire others to push their limits because I truly believe that we’re stronger than what we think. And I’ve been so pleasantly surprised that my body has held up. everything has held up and I’ve made it to 104.”She actually did not pick up the sport until after she lost the lower part of her left leg in 2001 to a type of cancer called Ewing sarcoma. Runners using a prosthetic leg can’t use a regular one. They need a running blade (刀锋式跑步义肢). Hunt-Broersma has two, but she was only able to run with one because of the swelling she experienced in her left leg that made it impossible to connect to the other blade. But even with the pain that occurred, she has been able to put up with it and continued.All together, Hunt-Broersma ran over 4,385 kilometers. During that time she has had a lot of time to think and has learned a few things about herself. “It has taught me how strong one can be and how important absolute determination is. If you’re mentally strong, you can do anything,” she said. “And our bodies are just amazing... This whole journey was impressive and super hard,but it’s told me how strong I can be as a person and how far I can push myself.”4.Why did Hunt-Broersma decide to run marathons?A.She wanted to set a new record.B.She was driven by her interestC.She intended to challenge herself.D.She was inspired by people like her 5.What difficulty did Hunt-Broersma meet with?A.She didn’t get professional training.B.She found it hard to keep her balanceC.She suffered discomfort in her left leg D.She couldn’t use regular running blades 6.What message does the author express through the story?A.Rome wasn’t built in a day.B.Time works wonders.C.A good beginning is half the battle.D.Nothing is impossible to a willing heart. 7.What may be the author’s attitude towards Hunt-Brocrsma?A.Annoyed.B.Admiring.C.Sympathetic.D.Critical.Exciting new research suggests there could be living things in the clouds above Venus, whose orbit brings it closer to the Earth than any other planets. A study of Venus’ atmosphere shows a type of gas is in its atmosphere. The gas, on the Earth, is only produced by very small creatures. If such creatures are found on Venus, it might be the first evidence of life existing somewhere else in the universe.Scientists found the gas in thick clouds about 35 miles above Venus’ equator. On the Earth, the gas is produced by special forms of life that live in airless environments such as swamps (沼泽). It can also be made in factories but neither explanation helps scientists to understand what the gas is doing on Venus.To get to the bottom of the mystery, the scientists have now made their data public in the hope that others may be able to find a clue that they have missed. Professor Greaves explained that the gas might be produced by some non-living natural process that we haven’t yet discovered, but for now, the idea of life on Venus has to be considered. She added, “In my whole career, I have been interested in the search for life elsewhere in the universe, so I’m just blown away that this is even possible. We are sincerely encouraging other people to tell us what we might have missed. This is how science works.” If scientists on the Earth are unable to find an explanation, then a future mission to Venus might be needed to prove whether or not life exists there. The US spaceagency NASA is already planning an exploration that they would float on a balloon through the clouds of Venus, and this could be launched sometime in the 2030s.8.What can we know about Venus according to Paragraph 1?A.The thickness of its clouds.B.The existence of a kind of gas.C.The distance from it to the Earth.D.The exact temperature of its atmosphere.9.Why have the scientists made their data public?A.To seek others’ help to find the truth.B.To change the wrong conclusion before.C.To prove the existence of life on Venus.D.To announce their successful experience.10.What does the underlined phrase “blown away” in Paragraph 3 probably mean?A.Impressed.B.Upset.C.Grateful.D.Doubtful. 11.Where does the text probably come from?A.A travel brochure.B.A medical report.C.A science magazine.D.A museum handbook.Paris embraces the Olympic spirit with a burst of creativity, as nearly 30 colorful street art pieces have transformed the city into an open- air gallery. These artworks, capturing scenes such as the dynamic French fencer Ysaora Thibus and the calm boaters on the Seine River, have made a lasting impression on people.Street artists come from different countries. Their pieces not only capture the essence of Parisian life, from the busy coffee shops to the charming bookstores, but also extend to airports in Lille, Lyon, and Marseille. These cities are co-hosting some of the Olympic events. Street art bridges the gap between high culture and everyday life, making art accessible to everyone. It promotes shared appreciation, inspiring conversations in public spaces like subway stations and airports and encouraging interaction among people of different backgrounds.The Olympics follow strict guidelines and show national spirit, while street art is freer and shows many different cultures from the community. Despite their differences, both the Olympicspirit and the world of art share a common platform of values that celebrate the pursuit of excellence, encouraging individuals to reach their highest potential. Creativity and innovation are central to both, driving artists to create with every brushstroke and athletes to excel in every performance as they both explore new heights of achievement.Art competitions first appeared at the Olympics in 1912 in Stockholm, with medals awarded in five categories: architecture, literature, music, painting and sculpture. However, the International Olympic Committee ended the competitions in the 1948 Games. Now, the spirit of the Olympics is reflected in the street art that brightens every corner of Paris. “Artists are like athletes. They also require a lot of effort and determination,” said JonOne, who is a pioneer of street art. “I respect athletes in basketball and runners. Art is not really a sport, but it should be included in the Olympics. Just surviving as an artist is an Olympic sport.”12.What do we know about street art according to paragraph 2?A.It centers around Olympic events.B.It helps strengthen social connections.C.It encourages cooperation between artists.D.It is displayed in open- air galleries in Paris.13.What do the Olympics and street art have in common?A.They have strict standards to follow.B.They reflect the creativity of communities.C.They emphasize the importance of competition.D.They inspire individuals to pursue their full potential.14.Which of the following statements will JonOne probably agree with?A.Art should be integrated with the Olympic Games.B.Artists deserve respect from people from all walks of life.C.Street artists should make use of the Olympics to gain popularity.D.The Olympics motivates many artists to create pioneering works.15.What is the best title for the text?A.Street Art Honors Olympic Athletes B.Street Art Adds Colors to the OlympicsC.The Olympics Draw Global Attention to Artists D.The Olympics Make Art Walk into People’s LifeA new urban sport, parkour (跑酷), is hitting the streets. It has developed from obstacle course training into a fitness option for young people. In parkour, the outside world is the gym!16 The goal of parkour is a direct route from one place to another. You meet an obstacle, you overcome it.Mark Toorock, who teaches the techniques of parkour at his fitness gym, says that parkour is a method to train the body and mind using obstacles as the medium. He says that this new sport is demanding and takes years to master. 17But Toorock, who used to be a martial arts expert, says that everyone can benefit from learning the basic skills like running, jumping and crawling. These are the things that humans used to have to do all the time. 18 The original idea of parkour was to return to running and jumping in moving from one place to another.Georges Hebert, a French military officer, was so impressed by the athletic performance of African tribes that he designed a training method based on running, climbing, jumping, balancing and throwing. 19 The French word for people who participate in the sport is ‘traceurs’ or ‘traceuses’.Dr Kenneth Kao explains that the sport of parkour is not extreme – it is the environment which is extreme and dangerous. Being outside, jumping off railings and flipping over park benches can be quite frightening, so parkour courses in gyms concentrate on practicing all the moves to make everything easier. 20 However, that is not real parkour because it’s indoors with a fixed obstacle. The goal for everyone is to go outdoors.A.Parkour was introduced into China in recent years and has gained popularity. B.Beginners should realize that they won’t be jumping over buildings any time soon!C.The bridges, buildings and railings (栏杆) of each and every city are the equipment. D.Every action in parkour is natural, so everyone must have the ability to move in this way. E.But today, due to modern transport, these basic skills are no longer used on a regular basis. F.Gyms provide thick floor matting (垫子) for rolling and rubberized boxes for jumping over. G.The word parkour comes from ‘parcours de combatant’, the French term for a military obstacle course.二、完形填空During the first weeks of his retirement from the fire service, Max had felt lost. And that’s when Polly, a library manager, helped him sign up as a 21 , for Robot Club in the library.Max hadn’t volunteered before and he didn’t know anything about 22 . He opened the packs containing the robots, nervously staring at the mess of parts. A dozen primary- age children gathered around the table, 23 eagerly. Almost double the number of adults crowded around them, watching.Glancing around the 24 at the many parents and carers accompanying the robot builders, Max knew he had to face this 25 .“Okay, team,” he started 26 , adopting the tone he’d used to 27 his unit during his firefighting days. “We’re going to build these robots one step 28 .”The children nodded, their hands busy with screws (螺丝) and plastic parts. Max wandered from table to table, 29 words of encouragement, and slowly but surely, 30 the instructions in his mind.An hour passed, and the room was filled with the sound of chatter and clinking (叮当作响) parts. As the final minutes of Robot Club ticked away, Max stood back and 31 the room. Robots of all shapes and sizes stood on the tables. The children were excited, 32 their creations to anyone who would look.As they began to leave, many children stopped to thank Max. Each “thank you” and “see you next week” gradually drove away the sense of 33 he had felt since retiring. He hadn’t put out a fire today, but perhaps he had 34 something far more important —a new 35 for himself.21.A.volunteer B.guard C.builder D.fireman 22.A.books B.programs C.children D.robots 23.A.searching B.waiting C.reading D.arguing 24.A.school B.neighbourhood C.team D.library 25.A.competition B.chance C.challenge D.conflict 26.A.calmly B.coldly C.keenly D.wildly 27.A.form B.brief C.warn D.scan28.A.at one time B.in no time C.at a time D.at no time 29.A.processing B.offering C.exchanging D.understanding 30.A.putting aside B.turning down C.keeping off D.piecing together 31.A.left B.surveyed C.decorated D.cleaned 32.A.bringing up B.calling on C.showing off D.giving away 33.A.freedom B.relief C.loss D.achievement 34.A.reviewed B.lighted C.promised D.preserved 35.A.skill B.belief C.approach D.purpose三、语法填空阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式。

重庆市外国语学校(四川外国语大学附属外国语学校)高一化学上学期半期考试试题

重庆市外国语学校(四川外国语大学附属外国语学校)高一化学上学期半期考试试题

重庆外国语学校2019-2019 学年(上)期中考试高一化学试题可能用到的相对原子质量:H-1C-12N-14O-16Na-23S-32Cl-35.5Ca-40 Ba-137Ⅰ卷一、选择题(共 16小题,每题 3分,共 48分,每题只有一个选项切合题意)1.化学与生活、社会发展息息有关,以下说法不正确的选项是( ) A.“熬胆矾铁釜,久之亦化为铜”,该过程发生了置换反响B.“霾尘聚集难见路人”,雾霾所形成的气溶胶有丁达尔效应C.“春蚕到死丝方尽,蜡炬成灰泪始干”诗句中波及氧化复原反响D.“青蒿一握,以水二升渍,绞取汁”,诗句中表现的屠呦呦对青蒿素的提取属于化学变化2.以下实验装置或操作与粒子直径的大小无直接关系的是()3.以下应用或事实与胶体的性质没有关系的是()....(3)用石膏或盐卤点制豆腐(4)尿毒症患者做“血液透析”(5)在河流入海口处易形成三角沙洲D.饱和氯化铁溶液中逐滴加入NaOH 溶液,产生红褐色积淀4.以下物质中既不可以导电,又不属于电解质的一组物质是()A.液氮、碳酸钙粉末B.干冰、金刚石C.碳酸溶液、食盐晶体D.KOH 溶液、熔融状态的NaCl5.以下反响中,水既不是氧化剂、也不是复原剂的氧化复原反响是()A.2F2+2H2O===4HF+O2B.Na2O+H2O===2NaOHC.2Na2O2+2H2O===4NaOH+O2↑D.Mg+2H 2O===Mg(OH) 2+H2↑6.以下物质的分类结果所有正确的选项是()A.矿泉水—混淆物胆矾—纯净物臭氧—单质B.纯碱—碱CH4—酸小苏打—酸式盐C.熟石灰—盐碳酸—含氧酸NH4Cl—酸D.液氧—非电解质硫酸钡—电解质NH3·H2O—弱电解质7.以下说法正确的选项是 ()A.分解反响必定是氧化复原反响B.SO2的水溶液能够导电,因此SO2是电解质C.氧化复原反响的本质是化合价的变化D.在反响 2H2S + SO2 = 3S↓+ 2HO 中,氧化产物和复原产物的物质的量之比为 2∶18.设 N A代表阿伏加德罗常数,以下说法中正确的选项是() A.2.3 克金属 Na 变为 Na+时失掉的电子数量为0.1N AB.22.4L 氢气所含原子数量为2N AC.0.1 mol NH4+中含有 0.1N A个电子D.常温下, 0.1mol/L MgCl 2溶液中含 Cl-数为 0.2N A9.以下离子方程式正确的选项是()A.醋酸与氢氧化钠溶液反响:H++OH-=H2OB.铜片插入硝酸银溶液:Cu + Ag+ =Cu2++AgC.碳酸钙与硝酸反响: CaCO+2H+=Ca2++H O+CO↑D.氢氧化钡溶液与稀H2SO4反响: Ba2++H++OH-+SO42-=BaSO4↓+H2O10.亚硝酸钠( NaNO2)可将人体血红蛋白中的Fe2+离子转变为 Fe3+离子,生成高铁血红蛋白而丧失与氧气联合的能力。

四川外语学院附属外国语学校(重庆外国语学校)简介

四川外语学院附属外国语学校(重庆外国语学校)简介

四川外语学院附属外国语学校(重庆外国语学校)简介

【期刊名称】《中考金刊》
【年(卷),期】2006(000)004
【摘要】四川外语学院附属外国语学校(重庆外国语学校)是1963年周恩来总
理亲自倡导下在全国首批创办的七所外国语学校之一,是以外语为特色,文理并重,注重学生综合素质和全面发展的重庆市教委直属重点中学,是重庆市唯一一所教育部认定的享有20%高校保送生资格的全国十三所学校之一。

【总页数】1页(PF0002)
【作者】无
【作者单位】无
【正文语种】中文
【中图分类】G639.28
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重庆市四川外国语大学附属外国语学校2022-2023学年八年级下学期开学考试英语试题(1)

重庆市四川外国语大学附属外国语学校2022-2023学年八年级下学期开学考试英语试题(1)

重庆实验外国语学校2022—2023学年度(下)初2024届开学定时作业英语试题(满分150分,120分钟完成)第Ⅰ卷(共77分)Ⅰ.听力测试。

(共20分)第一节(每小题1分,共6分)听一遍。

根据你所听到的句子,从A、B、C三个选项中选出最恰当的答语,并把答题卡上对应题目的答案标号涂黑。

1.A.I’m fine B.Not at all. C.The same to you.2.A.To the beach. B.Yes, I did. C.OK, let’s go.3.A.For seven days. B.By plane. C.Have a good time.4.A.Carrots. B.Cheese. C.Rice.5.A.I like French. B.It’s hard to read.C.By listening to CDs.6.A.It’s easy.B.Thank you. C.I have no idea.第二节(每小题1分,共6分)听一遍。

根据你所听到的对话和问题,从A、B、C 三个选项中选出正确答案,并把答题卡上对应题目的答案标号涂黑。

7.A.He had a fever. B.He had a toothache. C.He coughed a lot.8.A.In a bank. B.In a restaurant. C.In a hospital.9.A.A teacher. B.A nurse. C.A vet.10.A.At 2:30 p. m. B.At 3:00 p. m. C.At 3:30 p. m.11.A.Twice a day. B.Twice a week. C.Once a day.12.A.She’s having a rest. B.She’s cooking dinner.C.She’s enjoying a TV show.第三节(每小题1分,共4分)听两遍。

重庆市外国语学校四川外国语大学附属外国语学校2018_2019学年高二物理上学期半期考试试题

重庆市外国语学校四川外国语大学附属外国语学校2018_2019学年高二物理上学期半期考试试题

四川外国语大学附属外国语学校2018-2019学年度上学期半期考试高二物理试题一、选择题(本题共12小题,每小题4分.在每小题给出的四个选项中,第1-8题只有一项符合题目要求,第9-12题有多项符合题目要求.全部选对的得4分,选对但不全的得2分,有选错的得0分)1.关于电场强度有下列说法,正确的是()A.电场强度的方向总是跟电场力的方向一致B.在点电荷Q附近的任意一点,如果没有把试探电荷q放进去,则这一点的电场强度为零C.电场中某点的电场强度在数值上等于单位电荷在该点所受的电场力D.根据公式E=可知,电场强度跟电场力成正比,跟放入电场中的电荷的电量成反比2.关于电源的电动势,下列说法中正确的是( )A.电动势公式中W与电压中的W是一样的,都是电场力做的功B.不管外电路的电阻如何变化,电源的电动势不变C.电动势就是电压,它们之间无区别D.外电路全部断开时,电动势将变为零3.下列关于电阻率的叙述,正确的是( )A.电阻率与导体的长度和横截面积有关B.有些合金的电阻率几乎不受温度变化的影响,可以用来制成电阻温度计C.电阻率大的导体,电阻一定大D.电阻率表征了材料的导电能力的强弱,由导体的材料决定,且与温度有关4.如图所示,一质量为m、带电量为q的小球用细线系住,线的一端固在O点,若在空间加上匀强电场,平衡时线与竖直方向成60o角,则电场强度的最小值为()A.B.C.D.5.一电场的电场强度E随时间t变化的图象如图所示,此电场中有一个带电粒子,在t=0时刻由静止释放,若带电粒子只受电场力作用,则下列判断正确的是()A.带电粒子将向一个方向运动B.在1~3s内,电场力的功率为零C.3s末带电粒子的速度最大D.在2~4s内,电场力做的功等于零6.硅光电池是一种太阳能电池,具有低碳环保的优点.如图所示,图线a是该电池在某光照强度下路端电压U和电流I变化的关系图象(电池电动势不变,内阻不是定值),图线b是某电阻R的U–I图象.在该光照强度下将它们组成闭合回路时,下列说法中正确的是()A.硅光电池的内阻为6Ω B.硅光电池的总功率为0.4WC.硅光电池的内阻消耗的热功率为0.32WD.若将R换成阻值更大的电阻,硅光电池的输出功率增大7.在某个电场中,x轴上各点电势φ随x坐标变化如图所示,一质量m、电荷量+q的粒子只在电场力作用下能沿x轴做直线运动,下列说法中正确的是()A.x轴上x=x1和x=-x1两点电场强度和电势都相同B.粒子运动过程中,经过x=x1和x=-x1两点时速度一定相同C.粒子运动过程中,经过x=x1点的加速度大于x=x2点加速度D.若粒子在x=-x1点由静止释放,则粒子到达O点时刻加速度为零,速度达到最大8. 如图所示的电路,将两个相同的电流表分别改装成A1(0-3A)和A2(0-0.6A)的电流表,把两个电流表并联接入电路中测量电流强度,则下列说法正确的是()A.A2的指针还没半偏时,A1的指针已经半偏B.A1的指针还没半偏时,A2的指针已经半偏C.A1的读数为1A时,A2的读数为0.6AD.A1的读数为1A时,干路的电流I为1.2A9.如图所示的电路中,输入电压U恒为16V,灯泡L标有“6V,6W”字样,电动机线圈的电阻R M=1Ω,若灯泡恰能正常发光,则下列说法中正确的是()A.电动机的输入功率为16W B.电动机的输出功率为10WC.电动机的热功率为1W D.整个电路消耗的电功率为16W10.真空中有一竖直向上的匀强电场,其场强大小为E,电场中的A、B两点固定着两个等量异号点电荷+Q、−Q, A、B两点的连线水平,O为其连线的中点,c、d是两点电荷连线垂直平分线上的两点,Oc=Od, a、b两点在两点电荷的连线上,且Oa=Ob.下列判断正确的是( )A.a、b两点的电场强度相同B.c点的电势比d点的电势低C.将电子从a点移到c点的过程中,电场力对电子做正功D.将电子从a点移到b点时其电势能减小11.一平行板电容器的两块金属板A、B正对竖直放置,在金属板A的内侧表面系一绝缘细线,细线下端系一带电小球(可视为点电荷).两块金属板接在如图所示的电路中,电路中的R1为光敏电阻(其阻值随所受光照强度的增大而减小),R2 为滑动变阻器,R3为定值电阻.当R2的滑片P在中间时闭合电键S,此时电流表和电压表的示数分别为I和U,带电小球静止时绝缘细线与金属板A的夹角为θ.电源电动势E和内阻r一定,下列说法中正确的是()A.若将R2的滑动触头P向a端移动,则θ变小B.若将R2的滑动触头P向b端移动,则I减小,U增大C.保持滑动触头P不动,用较强的光照射R1,则小球重新达到稳定后θ变大D.保持滑动触头P不动,用较强的光照射R1,则平行板电容器的电量变小12.如图所示,a、b是带有同种电荷的小球,用绝缘细线挂于同一点,两球静止时离水平地面的高度相等,且不计空气阻力,两球带电荷量不变.若同时将两细线剪断,则下列说法中正确的是()A.a球先落地B.两球落地时动能相等C.落地时a球飞行的水平距离比b球小D.在飞行过程中,a球受到的冲量比b球受到的冲量大二、实验题(共26分)13.(12分)(1)如图所示游标卡尺的读数为______________cm(2)在测量金属丝电阻率的实验中,可供选用的器材如下:待测金属丝:R x(阻值约4 Ω,额定电流约0.5 A);电流表:A1 (量程3 A,内阻约0.05 Ω);A2(量程0.6 A,内阻约0.2 Ω);电压表:V1 (量程3 V,内阻约3 kΩ);V2(量程15 V,内阻约15 kΩ)电源:E(两节干电池);滑动变阻器:R(最大阻值约20 Ω);螺旋测微器;毫米刻度尺;开关S;导线.a. 用螺旋测微器测量金属丝的直径d,示数如图所示,读数为________mm.b.若滑动变阻器采用限流接法,为使测量尽量精确,电流表应选________、电压表应选________.c. 画线代替导线,将图中的电路连接完整.d . 如果用刻度尺测出被测金属丝的长度为l ,用螺旋测微器测出金属丝的直径为d ,用电流表和电压表分别测出通过金属丝的电流为I 和金属丝两端的电压为U .请用上述字母写出该金属丝电阻率的表达式为=____________.14.(8分)在“伏安法测电阻”的实验中:待测电阻R x 的阻值约为200 Ω,电压表V 的内阻约为2 k Ω,电流表A 的内阻约为10 Ω,测量电路中电流表的连接方式如图(a )或图(b )所示,结果由公式R x =U I计算得出,式中U 与I 分别为电压表和电流表的示数.若将图(a )和图(b )中电路测得的电阻值分别记为R x 1和R x2,则______(选填“R x 1”或“R x2”)更接近待测电阻的真实值,且测量值R x ________(选填“大于”“等于”或“小于”)真实值R 真,造成误差的原因是 ,若R A 越大,则越_______(填“大”或“小”).15. (6分)如图所示的是一种测量电容器电容的实验电路图,实验是通过对高阻值电阻放电,用电流传感器记录电流随时间的变化图线,进而测出电容器充电至电压U 时所带电荷量Q ,从而再求出待测电容器的电容C .某同学在一次实验时的情况如下: A 、按如图所示电路图接好电路;B 、接通开关S ,调节电阻箱R 的阻值,记下此时电流传感器的示数I 0=500μA ,电压表的示数U 0=8.0V ,I 0、U 0分别是电容器放电时的初始电流和电压;C 、断开开关S ,利用计算机和电流传感器,记录下电流随时间的变化关系,结果如图所示.(1)实验中电阻箱所接入阻值为R= Ω;(2)由上述i―t图像求出该电容器的放电总电荷量C(结果保留两位有效数字);(3)该电容器电容为F(结果保留两位有效数字).三、计算题(本大题3个小题,共36分,应写出必要的文字说明、方程式和重要的演算步骤,只写出最后答案的不得分,有数值计算的题,答案中必须明确写出数值和单位)16.(12分)如图a所示电路中,R1=4Ω,R2=6Ω,R3=3Ω.(1)若在C、D间连一个理想电压表,其读数为6V, 若在C、D间连一个理想电流表,其读数为1A,则电源的电动势E和内阻r分别为多大?(2)若在C、D间连一个阻值为3Ω的电阻R4,则电源的输出功率是多少?17. (12分)如图所示,光滑水平轨道与半径为R的光滑竖直半圆轨道在B点平滑连接,在过圆心O的水平界面MN的下方分布有水平向右的匀强电场.现有一质量为m、电荷量为+q的小球从水平轨道上A点由静止释放,小球运动到C点离开圆轨道后,经界面MN上的P点进入电场(P点恰好在A点的正上方,如图所示,小球可视为质点,小球运动到C点之前电荷量保持不变,经过C点后电荷量立即变为零).已知A、B间距离为2R,重力加速度为g,在上述运动过程中,求:(1)小球运动到C点的速度大小;(2)电场强度E的大小;(3)小球在圆轨道上运动时的最大速率;(4)小球对圆轨道的最大压力的大小.18. (12分)如图所示,在第一象限边长为L的正方形OABC区域内有一竖直向下的匀强电场,在第二象限直角边长为L的等腰直角三角形区域OAD内有一水平向右的匀强电场,两匀强电场场强E大小相等.一束带正电的粒子质量为m ,电荷量为q,以v0的初速从DF(DF垂直于x轴)沿x轴正向射入,从F点射入的粒子恰好从C点射出电场.(不计粒子的重力)(1)求两区域的场强大小E(2)若射入的粒子的纵坐标y=L/8,求该粒子打到x轴上的坐标(3)求粒子打在x轴上的范围四川外国语大学附属外国语学校2018-2019学年度上学期半期考试高二物理试题参考答案一:选择题二:实验题13. (1)R=1.6×104 Ω(2)41格 6.2×10-4C( 6.0×10-4C—— 6.3×10-4C内都对)(3) 7.8×10-5F (7.5×10-5F——7.9×10-5F)14: (1). 1.060 (2) 1.775 A2 V115:R x1大于电流表分压大三:计算题16:(1)若在C、D间连一个理想电压表,根据闭合电路欧姆定律,有理想电压表读数为U V=I1R2=6V若在C、D间连一个理想电流表,这时电阻R2与R3并联,并联电阻大小根据闭合电路欧姆定律,有理想电流表读数为(2)R总=r+R1+R并=9ΩI==P 输出=EI-I 2r=17:答案 (1)mg q (2)(2+22)gR (3)(2+32)mg解析 (1)设小球过C 点时速度大小为v C ,小球从A 到C 由动能定理知qE ·3R -mg ·2R =12mv 2C 小球离开C 点后做平抛运动到P 点,R =12gt 2 2R =v C t 得E =mg q(2)设小球运动到圆轨道D 点时速度最大,设最大速度为v ,此时OD 与竖直线OB 夹角设为α,小球从A 点运动到D 点的过程,根据动能定理知qE (2R +R sin α)-mgR (1-cos α)=12mv 2 即12mv 2=mgR (sin α+cos α+1) 根据数学知识可知,当α=45°时动能最大由此可得v =(2+22)gR(3)由于小球在D 点时速度最大且电场力与重力的合力恰好沿半径方向,故小球在D 点时对圆轨道的压力最大,设此压力大小为F ,由牛顿第三定律可知小球在D 点受到的轨道的弹力大小也为F ,在D 点对小球进行受力分析,并建立如图所示坐标系,由牛顿第二定律知,F -qE sin α-mg cos α=mv 2R解得F=(2+32)mg18(1)从F点发射的粒子在第二象限内做匀速直线运动,在第一象限电场区域内做类平抛运动,所以(2分)(2)在第二象限中,水平位移为,则应用动能定理(2分)若以此速度进入第一象限中的电场区域,则,说明粒子打在C点左侧(2分)(1分)(3)设在第二象限内的水平位移为x’,由动能定理有在第一象限内考虑飞出电场区域时仍未打在x轴上,则需同时满足:(3分)代入数据可解得当时,粒子飞出电场区域,飞出电场后在第一象限做匀速直线运动,设其匀速运动时间为t’,则(2分)可解得当时,x有最大值,其最大值为。

重庆市四川外国语大学附属外国语学校2024-2025学年九年级10月月考语文试题(含答案)

重庆市四川外国语大学附属外国语学校2024-2025学年九年级10月月考语文试题(含答案)

重庆实验外国语学校初2025届10月语文定时作业(满分150分,120分钟完成)一、基础知识及运用(30分)请阅读下面文段,按要求完成1~3题。

丝瓜、黄瓜是搭在架子上长的。

扁扁的绿叶在风中摇yè(),那些小黄花,就开在叶间,很妖娆地笑着。

南瓜多数是趴在地上长的,长长的蔓,会牵引得很远很远。

像对遥远的地方怀了无限向往,积攒着劲儿追寻了去,在一路的追寻中,绽放大朵大朵黄花,鲜妍()明媚。

最引人注目的,还是葵花。

它们挺立着,情绪饱满,斗志昂扬。

梵高在1888年的《向日葵》里,用大把金黄来渲染葵花。

梵高说,那是爱的最强光。

在困è()的日子里,那大朵的葵花,给他伤痕累累()的心,注入最后的温暖。

①沉甸甸的,恍若望得见日月风光在里头喧闹。

②葵花结籽,亦有另一种风韵。

③这个时候,它的头颅开始低垂,有些含羞,有些矫揉造作,但腰杆仍是挺直的。

④一颗瓜子,原来是一朵花的魂啊!⑤一颗一颗的瓜子,一日一日成形,饱满,吸足阳光和花香。

(——节选自丁立梅《祖母的葵花》,有删改)1.请为语段空缺处填写字音、字形。

(4分)(1)摇yè()(2)鲜妍()(3)困è()(4)伤痕累累()2.语段中加点的词语使用有误的一项是()(3分)A.妖娆B.积攒C.恍若D.矫揉造作3.画“_________”的句子语序排列最正确的一项是()(3分)A.②③①④⑤B.③⑤②④①C.②①③⑤④D.④⑤③①②4.语言陌生化是诗歌创作常用方法。

用新颖、形象的物象来表现日常、抽象的情感是路径之一,请从下面的备选词语里,任选一个创作一节小诗。

(4分)备选词语:忧伤愤怒欢笑示例——失落:假如时光已逝,鸟儿不再歌唱,风儿也吹倦了。

那就用黑暗的厚幕把我盖上,如同黄昏时节你用睡眠的衾被裹住大地,又轻轻合上睡莲的花瓣。

(——节选自《泰戈尔诗集》)5.名著阅读(8分)(1)下列对《艾青诗选》的表述有误的一项是()(4分)A.艾青诗歌的主要意象是“土地”和“太阳”。

重庆市外国语学校(四川外国语大学附属外国语学校)高一物理上学期半期考试试题

重庆市外国语学校(四川外国语大学附属外国语学校)高一物理上学期半期考试试题

重庆市外国语学校(四川外国语大学附属外国语学校)高一物理上学期半期考试试题〔考试时间:100分钟总分:100分〕一、选择题:此题共12小题,每题4分,共48分。

1~8题只要一个选项正确,9~12题有的有多个选项正确,全部选对的得4分,选对但不全的得2分,有选错的得0分。

1.以下说法正确的选项是:〔〕A. 质量很大的物体不能当作质点B. 物体的重心一定在物体上C. 绳的弹力只能沿绳收缩的方向D. 弹簧的弹力一定和弹簧的伸长成正比2.某同窗在水平空中上先向正南方向行走1.5km,再向正西方向行走2.0km,该同窗的位移大小和路程区分为:〔〕A.3.5km、3.5km B.2.5km、2.5km C.3.5km、2.5km D.2.5km、3.5km 3.以下说法中正确的选项是〔〕A.书放在水平桌面上遭到的支持力,是由于书发作了庞大形变而发生的B.用细木棍拨动浮在水中的圆木,圆木遭到的弹力是由于细木棍发作形变而发生的C.绳对物体的拉力方向总是沿着绳而指向绳拉长的方向D.杆对物体的弹力方向一定沿着杆的方向。

4.将物体竖直向上抛出后,能正确表示其速率v随时间t的变化关系的图线是图中〔〕A. B. C.D.5.在图中,假定两物体的重力区分是G A=20 N,G B=40 N,A,B间动摩擦因数μ1=0.2,B与空中之间动摩擦因数μ2=0.4,用力F作用在B上,使A、B匀速向右运动,A、B坚持相对运动,B对A的摩擦力为f A,空中对B的摩擦力为f B,那么〔〕A. f A=4N,方向向右B. f A=4N,方向向左 C.f B=24N,方向向左 D.f B=16N,方向向左6.四个图中,AB、BC均为轻质杆,各图中杆的A、C端都经过铰链与墙衔接,两杆都在B处由铰链衔接,且系统均处于运动形状.现用等长的轻绳来替代轻杆,能坚持平衡( )A.图中的AB杆可以用轻绳替代的有甲、乙、丙B.图中的AB杆可以用轻绳替代的有甲、丙、丁C.图中的BC杆可以用轻绳替代的有乙、丙、丁D.图中的BC杆可以用轻绳替代的有甲、乙、丁7.如下图,某学习小组应用直尺估测反响时间,将直尺刻度停止了改良,以相等时间距离在直尺上标志反响时间的刻度线,制造了〝反响时间测量仪〞,以下四幅图中刻度线标度正确的选项是〔〕A.B.C. D.8.一质点做匀减速直线运动,先后经过A、B、C三点,AB和BC两段位移大小之差为△x,质点经过两段位移的速度减大批相等,大小均为△v,那么质点的减速度大小为〔〕A. B. C. D.9. 以下说法正确的选项是〔〕A. 质点做竖直上抛运动,减速度方向先向上,后向下。

重庆市四川外国语大学附属外国语学校2022-2023学年高二上学期期末英语试题

重庆市四川外国语大学附属外国语学校2022-2023学年高二上学期期末英语试题

重庆外国语学校2022-2023学年度高2024届上期期末在线学习反馈检测英语试题(满分150分,120分钟完成)第Ⅰ卷选择题(共95分)第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。

录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A 、B 、C 三个选项中选出最佳选项。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirt?A.£19.15.B.£9.15.C.£9.18.答案是B 。

1.Why did the woman get a ticket?A.She parked in the wrong place.B.She drove too fast.C.She made a wrong turn.2.Who's Jenny?A.She is a student.B.She is a teacher.C.She is a doctor.3.Where are the two speakers?B.In a book store.A.In a classroom.4.What do we know about the womanC.In a post office.?A.She has been ill for a few months.B.She isn't worried about the exam.C.She is self-confident.5.Where did this conversation probably take place?A.In a restaurant.B.At a theatre.C.At a hospital.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

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重庆市外国语学校(四川外国语大学附属外国语学校)2018-2019学年高二数学上学期半期考试试题一.选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个备选项中,只有一项符合题目要求,把答案填写在答题卡相应位置上)1.集合{1,2,3,4,5,6}M =,集合{|36}N x N x =∈<<,则N M ⋂等于( ) A. {|46}x x << B.{|16}x x << C.{1,2,3,4,5,6} D. {4,5}2.已知函数220()30x x f x x x ⎧+≥⎪=⎨⎪+<⎩,则((1))f f -=( )A. 4B. 5C. 6D.7 3.下列函数中,在区间()0+∞,上是增函数的是( )A. 81y x =--B. 22y x =- C.1y x = D.12xy ⎛⎫= ⎪⎝⎭4.把(a -(1)a -移到根号内等于( )A.5.设30.8a =,0.83b =,3log 0.8c =,则,,a b c 的大小关系为( )A. c a b << (B )c b a << (C ) a b c << (D)a c b << 6.函数23log (87)y x x =-+-的单调递减区间为( )A. (,4)-∞B. (4,)+∞C. (1,4)D. (4,7) 7.若函数)10()(≠>=-a a a x f x且在),(+∞-∞上是减函数,则)1(log )(-=x x g a 的大致图象是( )8.已知f (x ),g (x )分别是定义在R 上的偶函数和奇函数,且f (x )-g (x )=x 3+x 2+2,则f (1)+g (1)=( )A .-2B .-1C .1D .29. 已知,1()(5)2,1x a x f x a x x ⎧>=⎨-+≤⎩,对任意21x x ≠,都有1212()()0f x f x x x ->-成立,则实数a的取值范围是( )A. 7(,5]2B.7[,5)2C. (1,5)D.(0,5]10.若函数(2)f x +为偶函数,且()y f x =在[2,)+∞为增函数,则下列结论正确的是( )A.(2)(3)(5)f f f -<<-B. (2)(5)(3)f f f -<-<C.(3)(5)(2)f f f <-<-D. (3)(2)(5)f f f <-<-11.已知函数x x f ln )(=,若b a <<0,且)()(b f a f =,则5a b +的取值范围是( )A.)+∞B.)+∞C.(6,)+∞D. [6,)+∞12.设函数1()2x f x x =-,1()2g x ax =-,若对任意的1[1,2]x ∈,都存在实数2[1,2]x ∈,使得()()12g x f x =成立,则实数a 的取值范围为( )A .[1,2]B .1[,3]2C .2[2,]5- D .3,22⎡⎤⎢⎥⎣⎦二、填空题(本大题4小题,每小题5分,共20分.把答案填写在答题卡相应位置上) 13.函数43x y a+=+ ()10≠>a a 的图象必过定点P ,P 点的坐标为_________.14. 已知函数)(x f 对任意实数b a ,,都有)()()(b f a f ab f +=成立,若(2)4f =,3)3(=f ,则)36(f 的值为___________.15.奇函数)(x f 对任意实数x 满足)()4(x f x f =+,且当)2,0[∈x ,()1f x x =-,则31()4f = . 16.已知函数21,1()4,1x f x x x x x ⎧-<-⎪=⎨⎪-+≥-⎩,如果方程()0f x a -=有三个不相等的实数解123,,x x x ,则123111x x x ++的取值范围 . 三、解答题:(17题10分,18,19,20,21,22题各12分。

解答时应写出必要的文字说明、演算步骤或推理过程,并答在答题卡相应的位置上.)17.(本小题满分10分,(1)小问6分,(2)小问4分,)(1) 已知13x x -+=.求22x x -+和1122x x -+的值.(2) 231lg162lg5log 3log 42++⋅18. (本小题满分12分)已知函数()f x A ,函数()2(1)x g x x m =-≤≤的值域为B .(1) 当1m =时,求A B ⋃;(2) 若()()R R A C B C B ⋃=,求实数m 的取指范围.19. (本小题满分12分)2()1ax f x bx =+是定义在R 上的函数,且12(1),(2)25f f ==. (1) 求,a b 的值,并判断函数)(x f 的奇偶性;(2) 利用函数单调性的定义证明:)(x f 在(1,1)-为增函数.20. (本小题满分12分)(1) 已知3(log )3,f x x =-求()f x 的解析式;(2) 当k 为何值时,方程|()|0f x k -=无解?有一解?有两解?21.(本小题满分12分)已知二次函数()2f x ax bx c =++(其中0a ≠)满足下列3个条件:①函数)(x f 的图象过坐标原点;②函数)(x f 的对称轴方程为21-=x ;③方程()f x x =有两个相等的实数根, (1) 求函数()f x 的解析式;(2) 令()()()12g x f x x λ=-+,若函数()g x 在[]2,1-上的最小值为3-,求实数λ的值.22. (本小题满分12分)已知函数2()21g x x x n =-++在[1,2]上有最大值1,设()()g x f x x=. (1) 求()f x 的解析式;(2) 若不等式22(log )2log 0f x k x -≥在[2,4]x ∈上有解,求实数k 的取值范围; (3) 若方程2(1)301x xkf e k e -+-=-有三个不同的实数解,求实数k 的取值范围.(e 为自然对数的底数).重庆外国语学校2018-2019学年第一学期高一年级期中考数学答案一、选择题二、填空题13. (4,4)- 14. 14 15. 3416.(3,)+∞ 三、解答题17.(1) 1122227;x xx x --+=+= (2)4 18. (1) 1[,)2A B ⋃=+∞ (2) [1,1)-19. (1) 解:1,1a b == 因为定义域为(-1,1), f(-x)=-=+-21xxf(x) ∴2()1xf x x =+是奇函数. ………6分 (2)设12,x x 为(-1,1)内任意两个实数,且12x x <,则22121221121212222222121212(1)(1)()(1)()()11(1)(1)(1)(1)x x x x x x x x x x f x f x x x x x x x +-+---=-==++++++ 又因为1211x x -<<<,所以12120,10x x x x -<->所以12()()0f x f x -<即12()()f x f x < 所以函数()f x 在(-1,1)上是增函数.…12分 20. (1)令3log x t=,得3tx =,所以()33t f t =-.所以()33xf x =-.………5分(2)0,k <无解0k =或者3k ≥时,有一解;03k << ,有两解; …………12分21. 解: (1)由题意得()00f =,即0c =. ………………………1分 ∵函数)(x f 的对称轴方程为21-=x ,∴122b a -=-,即a b =. …………………2分∴()2f x ax ax =+,∵方程()f x x =仅有一根,即方程()210ax a x +-=仅有一根,又0a ≠∴∆0=,即()210a -=,即1a =.∴()2f x x x =+. ……………………6分(2) ()22g x x x λ=-则函数()g x 的对称轴方程为x λ=①当2λ≤-时,函数()g x 在[]2,1-上单调递增.()min ()244g x g λ∴=-=+即443λ+=-,解得7=4λ-,故舍去. ……………………8分 ②当21λ-<<时,函数()g x 在[]2,λ-上单调递减,在[]1λ,上单调递增.()2min ()g x g λλ∴==-即23λ-=-,解得λλ== …………………1`0分 ③当1λ≥时,函数()g x 在[]2,1-上单调递减()min ()112g x g λ∴==-即123λ-=-,解得2λ=. ………………………11分综上:2λλ==. ………………………12分 22. (1) 2()(1)g x x n =-+()g x 在[1,2]上是增函数,所以(2)1g =,得0,n =1() 2.f x x x=+- ……………3分 (2) 由(1)得,1()2f x x x=+-,所以22(log )2log 0f x k x -≥等价于上有解,等价于2221221(log )log k x x ≤-+在在[2,4x ∈上有解,令211[,1]log 2t x =∈,则有2m a x 2(21)k t t ≤-+,所以124k ≤,所以k 得取值范围为1(,]8-∞. ……………………8分 (3)原方程可化为21(32)1(21)0xx e k e k --+-++=,令1xe q -=,则[0,)q ∈+∞.由题意得,2(32)(21)0q k q k -+++=有两个不同实数解,且1201,1q q <<≥.记2()(32)(21)h q q k q k =-+++,则(0)0(1)00h h >⎧⎪≤⎨⎪∆>⎩,解得0k >.所以实数k 的取值范围为(0,)+∞. ……………………12分重庆外国语学校备注:1.“考查知识点”为考查哪部分知识点,也可以注明用什么方法考查;2.“能力要求”为该知识点的课标要求,A---了解,B---理解,C----掌握;3.“预设难度”为在本试题中本知识点的预计难度系数,整卷难度系数为0.65左右. 其中选择题、填空题和解答题平均预设难度系数分别约为0.68,0.77和0.60.。

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